Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.lang.*;
public class Main{
public static void main(String[] args) throws FileNotFoundException, IOException{
        Scanner s = new Scanner(System.in);
        StringBuilder sb=new StringBuilder();
       int n=s.nextInt();
       double[] arr=new double[n+1];
       for(int i=1;i<=n;i++){arr[i]=s.nextDouble();}
        double[][] dp=new double[n+1][n+1];
      dp[0][0]=1;
      for(int i=1;i<=n;i++){
        dp[0][i]=dp[0][i-1]*(1-arr[i]);
      }
      for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
          if(i<=j){
            dp[i][j]=dp[i][j-1]*(1-arr[j])+dp[i-1][j-1]*arr[j];
          }
        }
      }
       double res=0;
       for(int i=n/2 +1;i<=n;i++){
        res+=dp[i][n];
       }
       System.out.println(res);
        }
    }